Questions Regarding this Chain Rule Proof

Question

I saw this proof of the Chain Rule on Hardy's A Course of Pure Mathematics (the notation I use will be a little different).

• Chain Rule:

Let $f\,\colon Y \subset \mathbb{R} \to \mathbb{R}$, $g\,\colon X \subset\mathbb{R} \to Y$, and $f \circ g\,\colon X \to \mathbb{R}$ be real-valued functions. Suppose that $f \circ g$ is differentiable at $x_0$, $f$ is differentiable at $g(x_0)$, and $g$ is differentiable at $x_0$. If $y = f(g(x))$ and $u = g(x)$, then $$\begin{align*} (f \circ g)'(x_0) &= (f' \circ g)(x_0) \cdot g'(x_0) \\[5pt] &\text{ or } \\[5pt] (f \circ g)'(x_0) &= f'(g(x_0)) \cdot g'(x_0). \\[5pt] &\text{ or } \\[5pt] \left. \frac{\mathrm{d}y}{\mathrm{d}x} \right|_{x=x_0} &= \left. \frac{\mathrm{d}y}{\mathrm{d}u} \right|_{u=g(x_0)} \cdot \left. \frac{\mathrm{d}u}{\mathrm{d}x} \right|_{x=x_0} \\[5pt] &\text{ or } \\[5pt] \left. \frac{\mathrm{d}f \circ g}{\mathrm{d}x} \right|_{x=x_0} &= \left. \frac{\mathrm{d}f}{\mathrm{d}u} \right|_{u=g(x_0)} \cdot \left. \frac{\mathrm{d}u}{\mathrm{d}x} \right|_{x=x_0} \end{align*}$$

• Proof:

Let $u_0 = g(x_0)$, $k = g(x_0 + h) - g(x_0)$, so that $k \to 0$ when $h \to 0$ and $$ \lim_{h \to 0} \frac{k}{h} = g'(x_0) \tag{i} $$ We must now distinguish two cases.

(a) Case 1:

Suppose that $g'(x_0) \neq 0$, and that $h$ is small but not zero. Then, $k \neq 0$ because of (i), and $$\begin{align*} (f \circ g)'(x_0) &= \lim_{h \to 0} \frac{f(g(x_0 + h)) - f(g(x_0))}{h} \\[5pt] &= \lim_{k \to 0} \frac{f(u_0 + k) - f(u_0)}{k} \cdot \lim_{h \to 0} \frac{k}{h} \\[5pt] &= f'(u_0) \cdot g'(x_0) = f'(g(x_0)) \cdot g'(x_0). \end{align*}$$

(b) Case 2:

Suppose that $g'(x_0) = 0$, and that $h$ is small but not zero. There are now two possibilities. If $k = 0$, then $$\begin{align*} (f \circ g)'(x_0) &= \lim_{h \to 0} \frac{f(g(x_0 + h)) - f(g(x_0))}{h} \\[5pt] &= \lim_{h \to 0} \frac{f(u_0 + k) - f(u_0)}{h} \\[5pt] &= \lim_{h \to 0} \frac{f(u_0) - f(u_0)}{h} \\[5pt] &= 0 = f'(g(x_0)) \cdot \underbrace{g'(x_0)}_{0}. \end{align*}$$

If $k \neq 0$, then $$\begin{align*} (f \circ g)'(x_0) &= \lim_{h \to 0} \frac{f(g(x_0 + h)) - f(g(x_0))}{h} \\[5pt] &= \lim_{k \to 0} \frac{f(g(x_0) + k) - f(g(x_0))}{k} \cdot \lim_{h \to 0} \frac{k}{h} \\[5pt] &= f'(g(x_0)) \cdot \underbrace{g'(x_0)}_{0} = 0. \tag*{$\blacksquare$} \end{align*}$$

• My Questions:

$1.$ In Case (2), why do we have to consider the case of $k \neq 0$ when $g'(x_0) = 0$?

I tried looking around and searching for functions where this does occur because I was confused on how $k$ can possibly be $\neq 0$ when $g'(x_0) = \lim\limits_{h \to 0} \frac{h}{k} = 0$.

I did find one example but it seems to be specifically a piecewise function ($x_0 = 0$):

$$\begin{align*} g(x) &= \begin{cases} x^2 \sin\left(\dfrac{1}{x}\right)& \text{if $x \neq 0$,} \\ 0 & \text{if $x = 0$.} \end{cases} \\[10pt] g'(0) &= \lim_{h \to 0} \frac{g(0 + h) - g(0)}{h} \\[10pt] &= \lim_{h \to 0} \frac{h^2\sin\left(\dfrac{1}{h}\right) - 0}{h} \\[10pt] &= \lim_{h \to 0} h \sin\left(\dfrac{1}{h}\right) \\ &= 0 \end{align*}$$

because $$\begin{align*} -1 &\leq \sin\left(\dfrac{1}{h}\right) \leq 1, \\[5pt] -\lim_{h \to 0}\, \lvert h \rvert &\leq \lim_{h \to 0} h \sin\left(\dfrac{1}{h}\right) \leq \lim_{h \to 0}\, \lvert h \rvert, \\[5pt] 0 &\leq \lim_{h \to 0} h \sin\left(\dfrac{1}{h}\right) \leq 0, \\[5pt] 0 &\leq g'(0) \leq 0. \tag{Squeeze Theorem} \end{align*}$$ Are there any examples of functions such that we need to consider proving the Chain Rule for the case of $k \neq 0$ but $g'(x_0) = 0$? Specifically, my question is do we really need to consider that case *if* we're strictly only considering proving the Chain Rule for non-piecewise differentiable functions.

$2.$ Is the Leibniz notation of the Chain Rule 'correct'? or can it be written more formally?

$3.$ Is this proof general and rigorous enough for differentiating single-variable function compositions? If not, can you show me a better and more rigorous proof, please (I haven't studied Real Analysis but I am somewhat comfortable with $\varepsilon-\delta$ proofs though I can't come up with them on my own).

Thank you.

Answer 1

1. $k$ is not $g'(x_0)$. $k = k(h) = g(x_0 + h) - g(x_0)$ is a function of $h$. Each value of $h$ gives a different value of $k$. While the limit as $h \to 0$ is $0$, this does not mean every value of $k$ has to be $0$. For example $x_0 = 0, g(x) = x^2$. Then $k(h) = (0+h)^2 - 0^2 = h^2$. If $h \ne 0$, then $k \ne 0$. Excepting when $g$ is constant near $x_0$, it is actually harder to come up with cases where $k = 0$ than where it does not.
2. Of course it is! Why do you suspect it isn't? As for writing it more formally, you have examples of exactly that in your post. What more are you looking for? There is nothing wrong with Leibnizian notation, provided you understand what it is actually saying and don't try to misinterpret it. Non-Standard Analysis even allows you to think of $\frac{dy}{dx}$ as being an actual division, though of infinitesimals, not ordinary numbers.
3. Not quite. The assumption that $f\circ g$ is differentiable at $x_0$ is unnecessary. It follows from the differentiability of $g$ at $x_0$ and $f$ at $g(x_0)$. They also make use of a few properties of limits that are presumably proved elsewhere. But then that is a main point of proving things: so that you can make use of them to simplify later proofs.